Unformatted text preview: 9 • More interesting than the instantaneous power is the timeaverage
power, which can be obtained from
Pav = 1
T T P (t)dt =
0 ω
2π 2π/ω P (t)dt (155) 0 • We can use the identity,
cos2 x = 1
(1 + cos 2x)
2 (156) And, remember that the integral of cos ωt over a period is 0,
that is, powers, yields,
T cos ωtdt = 0 (157) 0 • Expanding the incident power out,
P i (0, t) = V0+ 2
(1/2 + 1/2 cos(ωt + φ+ )) (W)
Z0 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (158) Electromagnetics I: Transmission lines so we have
i
Pav = 80 T 1
T P (t)dt =
0 V0+ 2
2Z0 (159) • The reﬂected power is found similarly, so we have the incident
and reﬂected average power (in Watts),
V0+ 2
(W)
2Z0 (160) V0+ 2
i
= −Γ2 Pav
2Z0 (161) i
Pav =
r
Pav = −Γ2 − r
Pav
= Γ2
i
Pav (162) • This is an important result: the ratio of the reﬂected and incident powers at the load give the reﬂection coeﬃcient magnitude
squared, or that the reﬂected power at the load is equal to the
incident power reduced by Γ2 term.
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 81 • The net average power delivered to the load is a sum of the incident and reﬂected powers:
i
r
Pav = Pav + Pav = V0+ 2
1 − Γ2 (W)
2Z0 (163) • Note, that we could have done the same thing (even easier) using
a phasor representation.
• The starting point for that is the average power relationship
which is,
1
˜˜
V · I∗
(164)
Pav =
2
• But, starting with that we still end up with,
Pav = V0+ 2
1 − Γ2 (W)
2Z0 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (165) Electromagnetics I: Transmission lines Zg 82 Transmission line
i
Pav +
~
Vg r
i
Pav = Γ2 Pav  ZL Figure 20: Power ﬂows on a transmission line. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Figure 219 Electromagnetics I: Transmission lines 83 1.10. Smith Chart
The Smith chart is a graphical tool for representing impedances, reﬂection coeﬃcients and transmission lines.
Let’s start with the polar representation of the reﬂection coeﬃcient, Γ.
Γ = Γejθr = Γ cos θr + j Γ sin θr = Γr + j Γi (166) • For passive load impedances, the reﬂection coeﬃcient is less then
unity, i.e. Γ ≤ 1.
• Impedances on the Smith chart are normalized (i.e. divided)
by a normalizing impedance. In our cases, that will be characteristic impedance Z0 of a transmission line, so that zL =
ZL /Z0 = rL + jxL .
• In that case we have,
Γ= 1+Γ
zL − 1
, or zL =
zL + 1
1−Γ Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (167) Electromagnetics I: Transmission lines 84
Gi 1 °
2
20 qr = G  = 1 D
1 0.9 0.7 0.5 0.3 0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.1 3
=5
qr q r = 180°
Shortcircuit
load q r = 90° A
G ° A = 0.5 C
0.1 0.3 B
0.2
G B = 0.54 0.3
0.4
0.5
0.6
...
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This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.
 Fall '12
 MartinSiderious
 Electromagnet

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